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This is a kind of a repost of a fascinating chess problem that appeared yesterday. The original poster got shy or something, and bafflingly deleted his own post, which was beginning to attract some good reputation and interest.

In the mean time, I have found some information which I want to share. So I will pose the question and then give one answer. Feel free to improve!

So: background. It seems that someone showed the original poster a problem a year ago, but the poster couldn't solve it, and wasn't quite sure if there was a solution at all. The problem is a selfmate: White to move has to coerce an unwilling Black into checkmating him. The poster didn't know how many moves it would take. Normally selfmates do have a specific move limit: they aren't like studies.

[Title "White to move and selfmate"]
[fen "kbQ5/pb6/P7/3B4/8/7p/7p/7K w - - 0 0"]

Is there a solution? If there is, what is it? And if there isn't, is the problem fixable?

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First the verdict on the posted problem. It has certainly no solution in up to 12 moves (checked by Popeye v4.79). And the author of the Gustav software which specializes in selfmates says that there is probably no regular solution to this problem.

Another expert thinks that if White simply captures everything which moves, he can then push bK over to g3 and force Black to win by h3xg2#. That's like taking a long chess problem and solving it like an regular chess game with no limit to the number of moves. Yes it will work to win but it's not really in the spirit of things. That's not to diminish his ingenious approach, but it's just not what we would consider a solution.

The good news is that this problem turns out to be a recognizable example of the Broecker schema, invented in 1891 by Gustav Von Broecker. The German PDB database contains 36(!) other problems which are all based on the same diagonal opposition of bishops. One early very clear example is the following:


    [Title "Broecker - London Chess Fortnightly 1892 - s#9"] 
    [fen "kb4R1/1b6/P7/8/8/7p/6BP/6BK w - - 0 0"] 

    1. Rf8 Bc6 2. Re8 Bd5 3. Rd8 Be4 4. Rc8 Bf3 5. Rh8 Be4 6. Bf3 Bd5 7. Be4 Bc6 8. Bd5 Bb7 9. Bc6 Bxc6#

"s#9" is shorthand for "selfmate in 9" - i.e. White moves first and forces an uncooperative Black to checkmate White by Black's 9th move at the latest.

As the original poster had deduced, there is a key relationship between the square of the queen (or rook) on the 8th rank, with the number of squares separating the bishops. Either player must always make them match if they can, and if he fails to, the other player gets a chance to make them match instead.

White wins because he is able to match by moving 1. Tf8 only. Black has many possible lines of play, but the longest he can hold out is 9 moves. One example solution is shown here.

In the original problem, the position is already a "match" with White to play. He is forced to break the match, and Black can always recreate it, so need never lose. It doesn't seem there is any way for White to use the additional power of the queen to change the parity.

However the matrix has proved very rich over the years, as you can see from the database, and I am sure there's more that can still be mined. The longest problem known to be sound is the following:


    [Title "Rotenburg & Trillon - Die Schwalbe 1980 - s#14"] 
    [fen "k7/1b4R1/N7/7p/8/7p/6BP/6BK w - - 0 0"] 

    1. Rf7! Bc6 2. Re7 Bd5 3. Rd7 Be4 4. Rc7 Bf3 5. Rh7 Be4 6. Bf3 Bd5 7. Be4 Bc6 8. Bd5 Bb7 9. Bc6 h4! 10. Bd5 Bc6 11. Rc7 Bb7 12. Rc1 Bc6 13. Rb1 Bb7 14. Bc6 Bxc6#

Again, here is a sample solution as Black has multiple choices, although White's move is unique at every point.

What's interesting about this one is that Black has a waiting move. This flips the parity, and so normally Black would be able to continue making matches, while White is breaking matches. However, White can flip the parity, by moving to the other side of the main diagonal. This can only be done while ensuring that b7 is never free for bK to enter. Either it must be occupied by bB, or attacked by wR.

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    "Another expert thinks that if White simply captures everything which moves, he can then push bK over to g3 and force Black to win by h3xg2#." - How would you even go about doing that? Capturing the b8 bishop seems impossible to do without just losing the queen, and capturing the light-squared bishop would be checkmate. – D M Apr 26 '18 at 18:53
  • Thanks for your response: here’s one idea. Start with 1. Dh8 Lc6 2. Dh6 & e.g. Ld6! 3. Lxc6+ Kb8 4. Dxd6+ Kc8 5. Dd7+ Kb8 6. Df7 Kc8 7. Dxa7, promote a rook on a8, or 2. ... e.g. Lb7! 3. Dc1 Lc6 4. Da1! Lb7 5. axb7+ and continue with the plan explained above. – Laska Apr 26 '18 at 20:26
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    In your second line, axb7 is checkmate? – D M Apr 26 '18 at 21:19
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    Try instead 1. Qh8 Bc6 2. Qa1 Bb7 3. Bc6 Be5 (best?) 4. axb7+ Kb8 5. Q~ (not 5. Qxe4#?) then force bKc7, promote wP=R, capture remaining bB push bK to g3, and then Rg2+ forcing hxg2#. – Laska Apr 27 '18 at 4:04
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Is there a solution?

I don't think so. If the White queen goes to a square where it can interpose, then Black can just play BxB.

If, for example, White tries Qe8-d7 or Qe8-e6, hoping Black will move his b8 bishop and allow Qc8+, Black can simply not move the b8 bishop. He could instead just move his light-squared bishop next to White's (White's queen cannot go from d7 or e6 to h8 to take advantage.)

And if there isn't, is the problem fixable?

The problem has a solution if White's queen or bishop start on another square. For example, starting the queen on d7 or g8 would provide lots of places for the queen to go, but only one is correct.

Another possibility is to add a white pawn on a4; moving the pawn loses a tempo and allows White to win. (Although that's not a satisfying way to construct the problem, since it has more than one solution; White could play either a4 or Qh8 as the first move.)

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